What is the integral from t to 3t of (sin(x))/(3x) ?

The integral from t to 3t of (sin(x))/(3x) is 1/3 (\Si(3t)-\Si(t))

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integral from t to 3t of (sin(x))/(3x)

\int_{t}^{3 t} \frac{sin ( x )}{3 x} d x

\frac{1}{3} (Si (3 t) - Si (t))

Solution steps

\int_{t}^{3 t} \frac{sin ( x )}{3 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{t}^{3 t} \frac{sin ( x )}{x} d x

Use the nonelementary integral:

\int \frac{sin ( x )}{x} d x = Si (x)

= \frac{1}{3} [Si (x)]_{t}^{3 t}

Compute the boundaries:

Si (3 t) - Si (t)

= \frac{1}{3} (Si (3 t) - Si (t))

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\int_{\frac{π}{3}}^{π} ∣ cos (x) ∣ d x

\int_{0}^{1} 3 x^{2} - 1 d x

\int_{0}^{0.3} ln (1 + x^{2}) d x

What is the integral from t to 3t of (sin(x))/(3x) ?
The integral from t to 3t of (sin(x))/(3x) is 1/3 (\Si(3t)-\Si(t))